Let $E/K$ be a field extension, $a,b\in E$ algebraic over $K$. Show: $\text{Min}(a,K,X)$ irreducible over $K(b)$ if and only if $\text{Min}(b,K,X)$ irreducible over $K(a)$

My attemp:

(1) Let deg $\text{Min}(a,K,X)$ = n, deg $\text{Min}(b,K,X)$ = m. We know: $[F(ab):F] = [F(ab):F(a)]\cdot [F(a):F] = [F(ab):F(a)]\cdot n $ $[F(ab):F] = [F(ab):F(b)]\cdot [F(b):F] = [F(ab):F(b)]\cdot m $

$\Rightarrow [F(ab):F(a)]\cdot n = [F(ab):F(b)]\cdot m$

Let $f:= \text{Min}(a,K,X)$ irreducible over $K(b)$. Since f is minimal Polynomial of $a$ over $K[X]$, it is the unique normalized irreducible polynomial with $f(a)=0$. Let $f'(a)=0$ and $f'(X)\in F(b)[X]$ be the minimal polynomial of $a$ over $K(b)$. Since $f\in K[X]$, also $f\in K(b)[X]$ and $\text{deg}\, f'\leq \text{deg}\, f$. Because $f:= \text{Min}(a,K,X)$ irreducible over $K(b) \Rightarrow f'=f \Rightarrow \text{deg}\, f'= \text{deg}\, f\Rightarrow [F(ab):F(b)]=n $.

Is this ok?


Your proof appears OK, except that

  • you switch from $K$ to $F$ and back (that's of course no serious problem)
  • you want to write $F(a, b)$ (or $K(a, b) $) instead of $F(ab) $

Also note that if you use the arguments of this answer, you'll see that your conditions are both equivalent to the symmetric condition $K(a) \cap K(b) = K$.


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